Solve the differential equation (y')^2+y^2-1=0

Solve the differential equation $(y')^2+y^2-1=0$.

Explanation of Method

Solve for $y'$ and separate, requiring $|y|\le1$; integrate using the standard $\arcsin$ integral, and check the boundary values $y=\pm1$ for singular solutions.

Step 1: Solve for $y'$ and separate

\[(y')^2=1-y^2 \ \Longrightarrow\ \frac{dy}{\sqrt{1-y^2}}=\pm dx,\qquad |y|\le1.\]

Step 2: Integrate

\[\arcsin y=\pm x+C \quad\Longrightarrow\quad y=\sin(x+c).\]

Step 3: Singular solutions

Setting $y'=0$ forces $y=\pm1$, and both satisfy $(0)^2+1-1=0$. These correspond to the maximum/minimum envelopes of the sine family $y=\sin(x+c)$.

Result

Final answer:

\[\boxed{\,y=\sin(x+c)\ \ \text{(general)}\qquad\text{and}\qquad y=\pm1\ \ \text{(singular)}.}\]